Completeness condition for Fréchet space

In Wikipedia's definition of Fréchet space, it is stated that a Fréchet space is a topological vector space that satisfies the following:

• It is locally convex
• Its topology can be induced by a translation-invariant metric
• Any (hence every) translation-invariant metric inducing the topology is complete

1) My first question lies with the "hence every" clause. It is known that completeness is a property of the metric, not topology, the standard example being $$(0,1)$$ under the absolute value and arctan metrics. As such how is it that we can conclude completeness here for every translation-invariant metric inducing the topology?

2) Wikipedia gives yet another equivalent condition for the completeness property - namely, let $$\{p_k\}_{k\in\mathbb{N}}$$ denote a countable family of seminorms defining the topology of $$X$$; then the space is complete with respect to the family of seminorms (i.e. if $$(x_n)$$ is a sequence in $$X$$ which is Cauchy with respect to each seminorm $$p_k$$, then there exists $$x\in X$$ such that $$(x_n)$$ converges to $$x$$ with respect to each seminorm $$p_k$$). How does the completeness with respect to any translation-invariant metric imply the completeness with respect to any countable family of seminorms and vice versa?

Let $$F$$ be your Fréchet space.

Suppose that $$d$$ is a translation invariant metric that induced the topology of $$F$$ and that $$F$$ is complete under $$d$$. Let $$\mathcal P=\{p_k\}_{k\in\mathbb N}$$ be any family of seminorms inducing the topology of $$F$$. Let $$d'(x,y)=\sum_{k=1}^\infty 2^{-k}\,\frac {p_k(x-y)}{1+p_k (x-y)}$$ be the metric induced by $$\mathcal P$$. It is easy to check that this metric induces the same topology that $$\mathcal P$$, that is the topology of $$F$$.

Looking at the respective unit balls, we can find (since both metrics induce the same topology) $$\alpha,\beta>0$$ such that $$\alpha\,d(x,0)\leq d'(x,0)\leq \beta\,d(x,0).$$ Because the metrics are translation invariant, for any $$x,y\in F$$ we have $$d(x,y)=d(x-y,0)$$ and the same for $$d'$$ so $$\tag1\alpha\,d(x,y)\leq d'(x,y)\leq \beta\,d(x,y)$$for all $$x,y\in F$$. Thus a sequence will be complete for $$d$$ if and only if it is complete for $$d'$$.

As $$(1)$$ is an equivalence relation, we have that completeness is the same for any translation invariant metric that induces the topology for $$F$$, and also agrees with completeness with respect to any sequence of seminorms that induces the topology.

• So in essence the argument is: every pair of equivalent translation-invariant metrics on a linear space is strongly equivalent (in the sense (1)) and strong equivalence of metrics preserves completeness (as opposed to simple topological equivalence). – Henno Brandsma Mar 3 '19 at 7:03
• Yes, if by "equivalent" you mean that the two metrics determine the same open sets around $0$ (or, what is the same, they generate the same topology). – Martin Argerami Mar 3 '19 at 12:14
• yes, that’s what I call topological equivalence. – Henno Brandsma Mar 3 '19 at 12:15
• $d'$ isn't a metric (the series may diverge to $\infty$). The usual modification is $d'(x,y)= \sum\limits_{k=1}^\infty 2^{-k} \frac{p_k(x-y)}{1+p_k(x-y)}$. – Jochen Mar 4 '19 at 12:36
• You are totally right. Editing right now. – Martin Argerami Mar 4 '19 at 13:04